On the Boundedness of the Hilbert Transform from Weighted Sobolev Space to Weighted Lebesgue Space

Abstract

Under some restrictions on weight functions we obtain sufficient conditions for the boundedness of the Hilbert transform from weighted Sobolev space of the first order on the semi-axis to weighted Lebesgue space.

1 Introduction

The important problem of harmonic analysis is to characterize weighted norm inequalities for the Hilbert transform. This problem has been studied since 1970s starting from the paper by Hunt–Muckenhoupt–Wheeden [2]. They proved that the Hilbert transform is bounded on weighted Lebesgue space $L^p_w$, $1<p<\infty$ (see definitions below), if and only if the weight w is from the Muckenhoupt $A_p$ class. Attempts to extend this result on the case of different weights and parameters of summation turned out to be difficult (see [3, 5, 6] for the case $p=2$ and different weights). However, some progress has been made for several restricted problems (see, for instance [8, 15]).

In this paper we study the mapping properties of Hilbert transform

$$\begin{aligned} Hf(x)= \,p.\,v.\,\int _0^\infty \frac{f(t)}{x-t}\, dt, \,\,x>0 \end{aligned}$$

between weighted Sobolev spaces of the first order and weighted Lebesgue spaces on the semi-axis.

Let $I:=(0,\infty )$, and let ${\mathfrak {M}}(I)$ be the set of all Lebesgue measurable functions on I,  ${\mathfrak {M}}^+(I)\subset {\mathfrak {M}}(I)$ be the subset of all non-negative functions, ${\mathfrak {M}}^\uparrow (I)\subset {\mathfrak {M}}^+(I)$ and ${\mathfrak {M}}^\downarrow (I)\subset {\mathfrak {M}}^+(I)$ be subsets of all non-negative, non-decreasing and non-increasing functions, respectively.

For $p\in (1,\infty )$ and $w\in {\mathfrak {M}}^+(I)$ we define

$$\begin{aligned} L^p_{w}(I):=\left\{ f\in {\mathfrak {M}}(I):\Vert f\Vert _{L^p_{w}}:=\left( \int _0^\infty |f|^pw^p\right) ^{1/p}<\infty \right\} . \end{aligned}$$

Let $v_0,v_1\in {\mathfrak {M}}^+(I)$, $v_1<\infty$ almost everywhere (a.e.) on I, $\frac{1}{v_1}\in L^{p'}_\text { loc}(I)$, $v_0\in L^p_\text { loc}(I)$, $\Vert v_0\Vert _{L^1(I)}>0$. Weighted Sobolev space is defined by

$$\begin{aligned} W^1_{p}(I):=\{u\in W^1_{1,\text { loc}}(I):\Vert u\Vert _{W^1_{p}}<\infty \}, \end{aligned}$$

where

$$\begin{aligned} \Vert u\Vert _{W^1_{p}}:=\Vert u\Vert _{L^{p}_{v_0}}+\Vert Du\Vert _{L^{p}_{v_1 }}. \end{aligned}$$

By AC(I) we denote the space of all functions absolutely continuous on every finite interval of I [7, Definition 3.1]. We also need subspaces where the second one is a completion in $W^1_p(I)$ of , which is the subspace of AC(I) such that

These Sobolev spaces were recently investigated in [9, 11,12,13,14]. In particular, a criteria for is given [9, Lemma 1.6], [12, (3.4)].

We study the problem of sufficient conditions for the boundedness of while $1<p,q<\infty$ and $v_0, v_1, w\in {\mathfrak {M}}^+(I).$ The weights $v_0$ and $v_1$ are chosen so that $\Vert f\Vert _{W^1_{p}}\approx \Vert Df\Vert _{L^p_{v_1}}.$

Section 2 contains some preliminaries.

The main results are in Sects. 3 and  4.

We use signs $:=$ and $=:$ for determining new objects, iff:= if and only if. We write $A\lesssim B,$ if $A\le cB$ with some positive constant c, which depends mainly on parameters of summation only. $A\approx B$ is equivalent to $A\lesssim B \lesssim A$. $\chi _E$ denotes the characteristic function (indicator) of a set E,  ${\mathbb {Z}}$ stands for the set of all integers. Uncertainties of the form $0\cdot \infty , \frac{\infty }{\infty }$ and $\frac{0}{0}$ are taken to be zero. $\Box$ stands for the end of a proof. If $1<p<\infty ,$ then $p':=\frac{p}{p-1}.$

2 Preliminaries

We suppose that weight functions $v_0$ and $v_1$ are such that

(2.1)

By [4, Remark 1.4] (2.1) is fulfilled iff

$$\begin{aligned} {{\mathbf {A}}}_0:=\sup _{(s,t)\subset (0,\infty )}\left( \int _s^t v_0^p\right) ^{1/p}\left( \min \left\{ \int _0^s v_1^{-p'}, \int _t^\infty v_1^{-p'}\right\} \right) ^{1/p'}<\infty . \end{aligned}$$

In this case

$$\begin{aligned} \Vert f\Vert _{W_{p}^1}\approx \Vert f'\Vert _{L^{p}_{v_1}}, \end{aligned}$$

(2.2)

that is the norm in two-weight Sobolev space $W_{p}^1(I)$ is equivalent to the norm $\Vert f'\Vert _{L^{p}_{v_1}}$ of “one-weight Sobolev space”. However, the first space is strictly smaller than the second, in general.

Lemma 2.1

Let $1<p<\infty$ and let the weights $v_0$ and $v_1$ be such that

$$\begin{aligned} {{\mathbf {A}}}_1:=\sup _{0<t<\infty }\left( \int _0^t v_0^p\right) ^{1/p}\left( \int _t^\infty v_1^{-p'}\right) ^{1/p'}<\infty . \end{aligned}$$

(2.3)

Then

(2.4)

where

$$\begin{aligned} Lg(x)&=\int _0^{x}\log \frac{x}{x-s}g(s)ds+\int _x^{2x}\log \frac{x}{s-x}g(s)ds\\&\quad +\int _{2x}^\infty \log \frac{x}{s-x}g(s)ds=:L_1g(x)+L_2g(x)-L_3g(x). \end{aligned}$$

Proof

Fix Since ${{\mathbf {A}}}_0\le {{\mathbf {A}}}_1,$ then (2.2) holds and, in particular,

$$\begin{aligned} \int _t^\infty |f'|\le \Vert f'\Vert _{L^{p}_{v_1}}\left( \int _t^\infty v_1^{-p'}\right) ^{1/p'},\,\,t>0. \end{aligned}$$

(2.5)

Then

$$\begin{aligned} Hf(x)= \,p.\,v.\,\int _0^\infty \frac{f(t)}{x-t}dt=\lim _{0<\varepsilon <x,\varepsilon \downarrow 0}\int _0^\infty \frac{\chi _{\varepsilon ,x}(t)f(t)}{x-t}dt, \end{aligned}$$

where

$$\begin{aligned} \chi _{\varepsilon ,x}(t):=\chi _{(0,x-\varepsilon )}(t)+\chi _{(x+\varepsilon ,\infty )}(t). \end{aligned}$$

Using (2.5) we write by Fubini’s theorem

$$\begin{aligned}&Hf(x)= \lim _{0<\varepsilon<x,\varepsilon \downarrow 0}\int _0^\infty \frac{\chi _{\varepsilon ,x}(t)\int _t^\infty f'(s)ds}{x-t}dt\\&=\lim _{0<\varepsilon<x,\varepsilon \downarrow 0}\int _0^\infty f'(s)\left( \int _0^s\frac{\chi _{\varepsilon ,x}(t)}{x-t}dt\right) ds=:\lim _{0<\varepsilon <x,\varepsilon \downarrow 0}\int _0^\infty k_{\varepsilon }(x,s)f'(s)ds. \end{aligned}$$

We see that $|k_{\varepsilon }(x,s)|$ is increasing with respect to $\varepsilon$ and

$$\begin{aligned} \lim _{0<\varepsilon <x,\varepsilon \downarrow 0} |k_{\varepsilon }(x,s)|=|k(x,s)|, \end{aligned}$$

where, $x\not = s,$

$$\begin{aligned} k(x,s)=\int _0^s\frac{dt}{x-t}={\left\{ \begin{array}{ll} \log \frac{x}{x-s}, &{}0<s<x,\\ \log \frac{x}{s-x}, &{}x<s<2x,\\ \log \frac{x}{s-x}, &{}s\ge 2x\end{array}\right. } \end{aligned}$$

(2.6)

and by the Lebesgue dominated theorem (2.4) follows. $\square$

Thus, by (2.2), (2.4), (2.6) and density of we have

and

(2.7)

where H is extended on by continuity.

In the next section we give sharp two-sided estimates for the norms $\Vert L_i\Vert _{L_{v_1}^p(I)\rightarrow L^q_{w}(I)}, i=1,2,3,$ which are of independent interest.

3 Auxiliary Results

The boundedness $\Vert L_1g\Vert _{L^q_w}\lesssim \Vert g\Vert _{L^p_{v_1}}$ is characterized in [1] under the condition $v_1\in {\mathfrak {M}}^\uparrow (I)$.

Theorem 3.1

[1] Let $1<p<\infty$, $0<q<\infty ,$ $1/r=(1/q-1/p)_+$, $v_1\in {\mathfrak {M}}^\uparrow (I).$ Then the inequality

$$\begin{aligned} \left( \int _0^\infty w^q(x)\left( \int _0^x \log \frac{x}{x-y}g (y)dy\right) ^qdx \right) ^{1/q}\le C\left( \int _0^\infty (gv_1)^p\right) ^{1/p},\,\,\,\,g\in {\mathfrak {M}}^+(I) \end{aligned}$$

holds iff

(i) $1<p\le q<\infty ,$ $A<\infty$, where

$$\begin{aligned} A:=\sup _{0<t<\infty }\left( \int _t^\infty \frac{w^q(x)dx}{x^q}\right) ^{1/q}\left( \int _0^t \frac{x^{p'}dx}{v_1^{p'}(x)}\right) ^{1/p'} \end{aligned}$$

(3.1)

and $C=\Vert L_1\Vert _{L_{v_1}^p(I)\rightarrow L^q_{w}}\approx A,$

(ii) $0<q<p<\infty , p>1,$ $B<\infty$, where

$$\begin{aligned} B:=\left( \int _0^\infty \left( \int _t^\infty \frac{w^q(x)dx}{x^q}\right) ^{r/q}\left( \int _0^t \frac{x^{p'}dx}{v_1^{p'}(x)}\right) ^{r/q'}\frac{t^{p'}}{v_1^{p'}(t)}dt\right) ^{1/r} \end{aligned}$$

(3.2)

and $C=\Vert L_1\Vert _{L_{v_1}^p(I)\rightarrow L^q_{w}}\approx B.$

The next theorem is devoted to the estimation of $L_2.$

Theorem 3.2

Let $1<p<\infty$, $1<q<\infty ,$ $1/r=(1/q-1/p)_+$, $v_1\in {\mathfrak {M}}^\uparrow (I)$ and there exists $\gamma >0$ such that $x^{-\gamma }v_1(x)\in {\mathfrak {M}}^\downarrow (I).$ Then the inequality

$$\begin{aligned} \left( \int _0^\infty w^q(x)\left( \int _x^{2x} \log \frac{x}{y-x}g (y)dy\right) ^qdx \right) ^{1/q}\le C\left( \int _0^\infty (gv_1)^p\right) ^{1/p},\,\,\,\,g\in {\mathfrak {M}}^+(I) \end{aligned}$$

(3.3)

is valid iff the inequality

$$\begin{aligned} \left( \int _0^\infty \frac{w^q(x)}{x^q}\left( \int _{\frac{3x}{2}}^{2x} (2x-y)g(y)dy\right) ^qdx \right) ^{1/q}\le D\left( \int _0^\infty (gv_1)^p\right) ^{1/p},\,\,\,\,g\in {\mathfrak {M}}^+(I) \end{aligned}$$

(3.4)

holds and ${\mathbb {A}}<\infty ,$ where

$$\begin{aligned} {\mathbb {A}}:={\left\{ \begin{array}{ll}\displaystyle \sup _{k\in {\mathbb {Z}}}\sup _{t\in [a^k,a^{k+1}]} \left( \int _{a^{k}}^t\omega ^q\right) ^{1/q}\left( \int _t^{a^{k+2}} u^{p'}\right) ^{1/p^\prime }, &{} p\le q,\\ \displaystyle \left( \sum _{k\in {\mathbb {Z}}}\int _{a^k}^{a^{k+1}}\omega ^q(x) \left( \int _{a^{k}}^x\omega ^q\right) ^{r/p}\left( \int _x^{a^{k+2}} u^{p'}\right) ^{r/p^\prime }dx\right) ^{1/r}, &{} q<p,\end{array}\right. } \end{aligned}$$

(3.5)

where $a\in (1,\sqrt{2}),$ $\omega (x):=\frac{w(x)}{x^{\gamma }},$ $u(y):=\frac{y^{\gamma }}{v_1(y)}.$ Moreover, $C=\Vert L_2\Vert _{L_{v_1}^p(I)\rightarrow L^q_{w}}\approx D+{\mathbb {A}}$.

Proof

The inequality (3.3) is equivalent to two inequalities

$$\begin{aligned} \left( \int _0^\infty w^q(x)\left( \int _x^{\frac{3x}{2}} \log \frac{x}{y-x}g (y)dy\right) ^qdx\right) ^{1/q}\le C_1\left( \int _0^\infty (gv_1)^p\right) ^{1/p},\,\,\,\,g\in {\mathfrak {M}}^+(I) \end{aligned}$$

(3.6)

and

$$\begin{aligned} \left( \int _0^\infty w^q(x)\left( \int _{\frac{3x}{2}}^{2x} \log \frac{x}{y-x}g (y)dy\right) ^qdx \right) ^{1/q}\le C_2\left( \int _0^\infty (gv_1)^p\right) ^{1/p},\,\,\,\,g\in {\mathfrak {M}}^+(I). \end{aligned}$$

(3.7)

Moreover,

$$\begin{aligned} C\approx C_1+C_2. \end{aligned}$$

If $y\in (3x/2,2x),$ then

$$\begin{aligned} \log \frac{x}{y-x}\approx \frac{2x-y}{x}. \end{aligned}$$

Therefore, (3.7) is equivalent to (3.4) and $C_2\approx D.$ Put $u(y):=\frac{y^{\gamma }}{v_1(y)},$ $\omega (x):=\frac{w(x)}{x^{\gamma }}.$ Then $u\in {\mathfrak {M}}^\uparrow (I)$ and (3.6) is equivalent to

$$\begin{aligned} \left( \int _0^\infty \omega ^q(x)\left( \int _x^{\frac{3x}{2}} \log \frac{x}{y-x}g (y)u(y)dy\right) ^qdx\right) ^{1/q}\le C_1\left( \int _0^\infty g^p\right) ^{1/p},\,\,\,\,g\in {\mathfrak {M}}^+(I). \end{aligned}$$

Let $a\in (1,\sqrt{2})$ and

$$\begin{aligned} T_{k,1}g(x):= & {} \omega (x)\chi _{[a^k,a^{k+1}]}(x)\int _x^{a^{k+1}}\log \frac{x}{s-x}g(s)u(s)ds,\\ T_{k,2}g(x):= & {} \omega (x)\chi _{[a^k,a^{k+1}]}(x)\int _{a^{k+1}}^{a^{k+2}}\log \frac{x}{s-x}g(s)u(s)ds,\\ T_k:= & {} T_{k,1}+T_{k,2},\,\,\,T_1:=\sum _{k\in {\mathbb {Z}}}T_{k,1},\,\,\, T_{2}:=\sum _{k\in {\mathbb {Z}}}T_{k,2},\,\,\,T:=T_{1}+T_{2}. \end{aligned}$$

The operators $T_1$ and $T_2$ are block-diagonal. Denote $\Vert S\Vert :=\Vert S\Vert _{L^p\rightarrow L^q}.$ Then by [16, Lemma 1]

$$\begin{aligned} \Vert T_1\Vert&=\sup _{k\in {\mathbb {Z}}}\Vert T_{k,1}\Vert , \,\,\Vert T_2\Vert =\sup _{k\in {\mathbb {Z}}}\Vert T_{k,2}\Vert , \, \, p\le q,\\&\Vert T_1\Vert =\left( \sum _{k\in {\mathbb {Z}}}\Vert T_{k,1}\Vert ^r\right) ^{1/r}, \,\, \Vert T_2\Vert =\left( \sum _{k\in {\mathbb {Z}}}\Vert T_{k,2}\Vert ^r\right) ^{1/r} \, \, q<p. \end{aligned}$$

If $p\le q,$ then

$$\begin{aligned} \Vert T\Vert \le \Vert T_1\Vert +\Vert T_2\Vert \le 2\sup _{k\in {\mathbb {Z}}}\Vert T_k\Vert \end{aligned}$$

and

$$\begin{aligned} 2\Vert T\Vert \ge \Vert T_1\Vert +\Vert T_2\Vert \ge \sup _{k\in {\mathbb {Z}}}(\Vert T_{k,1}\Vert +\Vert T_{k,2}\Vert )\ge \sup _{k\in {\mathbb {Z}}}\Vert T_k\Vert . \end{aligned}$$

If $q<p,$ then

$$\begin{aligned} \Vert T\Vert \le \Vert T_1\Vert +\Vert T_2\Vert \le 2\left( \sum _{k\in {\mathbb {Z}}}\Vert T_k\Vert ^r\right) ^{1/r} \end{aligned}$$

and

$$\begin{aligned} 2\Vert T\Vert \ge \Vert T_1\Vert +\Vert T_2\Vert \ge \left( \sum _{k\in {\mathbb {Z}}}\Vert T_{k,1}\Vert ^r\right) ^{1/r}+\left( \sum _{k\in {\mathbb {Z}}}\Vert T_{k,2}\Vert ^r\right) ^{1/r} > rsim \left( \sum _{k\in {\mathbb {Z}}}\Vert T_k\Vert ^r\right) ^{1/r}. \end{aligned}$$

For $x\in [a^k,a^{k+1}]$, $s\in [x,a^{k+2}]$

$$\begin{aligned} \frac{x}{s-x}\ge \frac{1}{a^2-1}>1. \end{aligned}$$

Hence, it follows from

$$\begin{aligned} \Vert T_kg\Vert _{L^q}\le \Vert T_k\Vert \Vert g\Vert _{L^p} \end{aligned}$$

that

$$\begin{aligned} \left( \int _{a^k}^{a^{k+1}}\omega ^q(x)\left( \int _x^{a^{k+2}}gu\right) ^qdx\right) ^{1/q}\lesssim \Vert T_k\Vert \Vert g\chi _{[a^k,a^{k+2}]}\Vert _{L^p}. \end{aligned}$$

(3.8)

Let $A_k$ be a constant characterising Hardy’s inequality (3.8), i.e.

$$\begin{aligned} A_k:={\left\{ \begin{array}{ll}\displaystyle \sup _{t\in [a^k,a^{k+1}]} \left( \int _{a^{k}}^t\omega _k^q\right) ^{1/q}\left( \int _t^{a^{k+2}} u^{p'}\right) ^{1/p^\prime }, &{} p\le q,\\ \displaystyle \left( \int _{a^k}^{a^{k+1}}\omega _k^q(x) \left( \int _{a^{k}}^x\omega _k^q\right) ^{r/p}\left( \int _x^{a^{k+2}} u^{p'}\right) ^{r/p^\prime }dx\right) ^{1/r}, &{} q<p,\end{array}\right. } \end{aligned}$$

where $\omega _k:=\omega \chi _{[a^k,a^{k+1}]}.$ Then $A_k\lesssim \Vert T_k\Vert$ and

$$\begin{aligned}&\left( \int _0^\infty \omega ^q(x)\left( \int _x^{\frac{3}{2}x}\ln \frac{x}{s-x}g(s)u(s)ds\right) ^qdx \right) ^{1/q}\le \Vert Tg\Vert _{L^q_\omega }\\&\le \left( \int _0^\infty \omega ^q(x)\left( \int _x^{2x}\ln \frac{x}{s-x}g(s)u(s)ds\right) ^qdx \right) ^{1/q}. \end{aligned}$$

Hence,

$$\begin{aligned} {\mathbb {A}}\lesssim \Vert T\Vert \lesssim C. \end{aligned}$$

Thus,

$$\begin{aligned} C_2+{\mathbb {A}}\lesssim C\lesssim C_2+\Vert T\Vert \end{aligned}$$

and it is sufficient to prove that

$$\begin{aligned} \Vert T\Vert \lesssim {\mathbb {A}}. \end{aligned}$$

Let $p\le q$. For $x\in [a^k,a^{k+1}]$ we show that

$$\begin{aligned}&I:=\int _x^{a^{k+2}}\log \frac{x}{y-x}g(y)u(y)dy\nonumber \\&\lesssim \left( \int _x^{a^{k+2}}g^p(y)\left[ \int _y^{a^{k+2}}u^{p^\prime }\right] ^{\frac{1}{p^\prime }}dy\right) ^{\frac{1}{p}}\left( \int _x^{a^{k+2}}u^{p^\prime }\right) ^{\frac{1}{{p^{\prime }}^2}}. \end{aligned}$$

(3.9)

First by Hölder’s inequality

$$\begin{aligned} I=&\int _x^{a^{k+2}}\left\{ g(y)\left[ \int _y^{a^{k+2}}u^{p^\prime }(t)\left[ \log \frac{x}{t-x}\right] ^{p'}dt\right] ^{\frac{1}{pp^\prime }}\right\} \\&\times \left\{ u(y)\log \frac{x}{y-x}\left( \int _y^{a^{k+2}}u^{p^\prime }(t)\left[ \log \frac{x}{t-x}\right] ^{p'}dt\right) ^{-\frac{1}{pp^\prime }}\right\} dy\le I_2\cdot I_1, \end{aligned}$$

where

$$\begin{aligned} I_2:=\left( \int _x^{a^{k+2}}g^p(y)\left( \int _y^{a^{k+2}}u^{p^\prime }(t)\left[ \log \frac{x}{t-x}\right] ^{p'}dt\right) ^{\frac{1}{p^\prime }}dy\right) ^{\frac{1}{p}} \end{aligned}$$

and

$$\begin{aligned} I_1:=&\left( \int _x^{a^{k+2}}u^{p^\prime }(y)\left[ \log \frac{x}{y-x}\right] ^{p^\prime } \left( \int _y^{a^{k+2}}u^{p^\prime }(t)\left[ \log \frac{x}{t-x}\right] ^{p^\prime }dt\right) ^{-\frac{1}{p}}dy \right) ^{\frac{1}{p^\prime }}\\&\approx \left( \int _x^{a^{k+2}}u^{p^\prime }(t)\left[ \log \frac{x}{t-x}\right] ^{p^\prime }dt \right) ^{\frac{1}{{p^\prime }^2}}=:I_3^{\frac{1}{{p^\prime }^2}}(x). \end{aligned}$$

Observe that for $x\ge 1$, $\sigma >0$

$$\begin{aligned} \log x=\int _1^x\frac{dt}{t}\le \int _1^x\frac{t^\sigma dt}{t}=\frac{1}{\sigma }x^\sigma , \end{aligned}$$

therefore, $\log \frac{x}{x-s}\le \frac{1}{\sigma }\left( \frac{x}{x-s}\right) ^\sigma .$ Applying the Chebyshev inequality: if $F\in {\mathfrak {M}}^\uparrow (a,b),$ $G\in {\mathfrak {M}}^\downarrow (a,b),$ then

$$\begin{aligned} \int _a^bFG\le \frac{1}{b-a}\left( \int _a^bF\right) \left( \int _a^bG\right) , \end{aligned}$$

for $u\in {\mathfrak {M}}^\uparrow (I)$ and $\log \frac{x}{\cdot -x}\in {\mathfrak {M}}^\downarrow (I)$ and taking $\sigma \in (0,1/p')$ we find

$$\begin{aligned} I_3(x,y):=&\int _y^{a^{k+2}} u^{p^\prime }(t)\left[ \log \frac{x}{t-x}\right] ^{p^\prime } dt \le \frac{1}{a^{k+2}-y}\int _y^{a^{k+2}}u^{p^\prime }\int _y^{a^{k+2}}\left[ \log \frac{x}{t-x}\right] ^{p^\prime }dt\\&\lesssim \frac{x^{\sigma p'}}{a^{k+2}-y} \int _y^{a^{k+2}}u^{p^\prime }\int _y^{a^{k+2}}\frac{dt}{(t-x)^{\sigma p'}}. \end{aligned}$$

By elementary inequality [10, p. 139]: if $b>a>0, \tau >0,$ then

$$\begin{aligned} b^\tau -a^\tau \approx (b-a)b^{\tau -1} \end{aligned}$$

we find

$$\begin{aligned} \int _y^{a^{k+2}}\frac{dt}{(t-x)^{\sigma p'}}\approx \left( a^{k+2}-x\right) ^{1-\sigma p'} -\left( y-x\right) ^{1-\sigma p'}\approx (a^{k+2}-y)\left( a^{k+2}-x\right) ^{-\sigma p'} \end{aligned}$$

and

$$\begin{aligned} \frac{x}{a^{k+2}-x}\le \frac{a^{k+1}}{a^{k+2}-a^k}=\frac{a}{a^2-1}\in (0,\infty ). \end{aligned}$$

Hence,

$$\begin{aligned} I_3(x,y)\lesssim \int _y^{a^{k+2}}u^{p^\prime }=I_3(y). \end{aligned}$$

Thus,

$$\begin{aligned}&I_1\lesssim \left( \int _y^{a^{k+2}}u^{p^\prime }\right) ^{\frac{1}{{p^\prime }^2}},\\&I_2\lesssim \left( \int _x^{a^{k+2}}g^p(y)\left( \int _y^{a^{k+2}}u^{p^\prime }\right) ^{\frac{1}{p^\prime }}dy\right) ^{\frac{1}{p}} \end{aligned}$$

and (3.9) follows. Now we write by (3.9) and Minkowskii’s inequality

$$\begin{aligned}&\Vert T_kg\Vert _{L^q}^q=\int _{a^k}^{a^{k+1}}\omega ^q(x)\left( \int _x^{a^{k+2}}\log \frac{x}{y-x}g(y)u(y)dy\right) ^qdx\\&\lesssim \int _{a^k}^{a^{k+1}}\omega _k^q(x)\left( \int _x^{a^{k+2}}g^p(y)\left( \int _y^{a^{k+2}}u^{p^\prime }\right) ^{\frac{1}{p^\prime }}dy\right) ^{\frac{q}{p}}\left( \int _x^{a^{k+2}}u^{p^\prime }\right) ^{\frac{q}{{p^\prime }^2}}dx\\&\le \left( \int _{a^k}^{a^{k+2}}g^p(y)\left( \int _y^{a^{k+2}}u^{p^\prime }\right) ^{\frac{1}{p^\prime }} \left( \int _{a^k}^y\omega _k^q(x)\left( \int _x^{a^{k+2}}u^{p^\prime }\right) ^ {\frac{q}{{p^\prime }^2}} dx\right) ^{\frac{p}{q}}dy\right) ^{\frac{q}{p}}. \end{aligned}$$

Since

$$\begin{aligned} \left( \int _x^{a^{k+2}}u^{p^\prime }\right) ^{\frac{1}{p^\prime }}\le A_k\left( \int _{a^k}^{\min \{x,a^{k+1}\}} \omega _k^q\right) ^{-\frac{1}{q}},\,\,x\in (a^k,a^{k+2}) \end{aligned}$$

then for $a^k<x<y<a^{k+2}$

$$\begin{aligned} \int _{a^k}^y\omega _k^q(x)\left( \int _x^{a^{k+2}}u^{p^\prime }\right) ^{\frac{q}{{p^\prime }^2}}&\le A_k^{\frac{q}{p'}}\int _{a^k}^y\omega _k^q(x)\left( \int _{a^k}^{\min \{x,a^{k+1}\}} \omega _k^q\right) ^{-\frac{1}{p^\prime }}dx\\&\approx A_k^{\frac{q}{p'}}\left( \int _{a^k}^{\min \{y,a^{k+1}\}} \omega _k^q\right) ^{\frac{1}{p}}. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert T_kg\Vert _{L^q}^q&\lesssim A_k^{\frac{q}{p'}} \left( \int _{a^k}^{a^{k+2}}g^p(y)\left( \int _y^{a^{k+2}}u^{p^\prime }\right) ^{\frac{1}{p^\prime }}\left( \int _{a^k}^y\omega _k^q\right) ^{\frac{1}{q}}dy\right) ^{\frac{q}{p}}\\&\lesssim A_k^q\Vert g\chi _{[a^k,a^{k+2}]}\Vert _{L^p} \end{aligned}$$

and $\Vert T\Vert \lesssim {\mathbb {A}}$ for $p\le q$ follows by Jensen’s inequality.

Now let $q<p.$ Put

$$\begin{aligned} h(x):=\chi _{[a^{k},a^{k+2}]}(x)\left( \int _x^{a^{k+2}}\left[ \int _s^{a^{k+2}}u^{p^\prime }\right] ^{\frac{r}{q^\prime }}\left[ \int {a^k}^s \omega ^q_k\right] ^{\frac{r}{p}}u(s)^{p^\prime }ds\right) ^{\frac{q}{r}}. \end{aligned}$$

We have by Hölder’s inequality with exponents p/q and r/q

$$\begin{aligned}&\left( \int _{a^{k}}^{a^{k+1}}\omega _k^q(x)\left( \int _x^{a^{k+2}}\ln \frac{x}{y-x}f(y)u(y)dy\right) ^qdx\right) ^{\frac{1}{q}} \\&\le \left( \int _{a_{k}}^{a^{k+1}}\omega _k(x)^q[h(x)]^{\frac{r}{q}}dx\right) ^{\frac{1}{r}}\times \\&\left( \int _{a^{k}}^{a^{k+1}}\omega _k^q(x)[h(x)]^{-\frac{p}{q}} \left( \int _x^{a^{k+2}}\log \frac{x}{y-x}f(y)u(y)dy\right) ^p dx\right) ^{\frac{1}{p}}. \end{aligned}$$

Integrating by parts we have

$$\begin{aligned}&[h(x)]^{\frac{r}{q}}= \int _x^{a^{k+2}} \left( \int _{a^{k}}^s \omega _k^q\right) ^{\frac{r}{p}} d\left( -\left( \int _s^{a^{k+2}}u^{p^\prime }\right) \right) ^{\frac{r}{p^\prime }}\\&\approx \left( \int _{a^{k}}^x\omega _k^q\right) ^{\frac{r}{p}} \left( \int _x^{a^{k+2}}u^{p^\prime }\right) ^{\frac{r}{p^\prime }} +\int _x^{a^{k+1}}\left( \int _s^{a^{k+2}}u^{p^\prime }\right) ^{\frac{r}{p^\prime }} \left( \int _{a^{k}}^s\omega _k^q\right) ^{\frac{r}{p}-1} \omega _k^q(s)ds. \end{aligned}$$

It implies

$$\begin{aligned} \left( \int _{a^k}^{a^{k+1}}\omega _k^qh^{\frac{r}{q}}\right) ^{1/r}\lesssim \left( \int _{a^k}^{a^{k+1}}\left( \int _s^{a^{k+2}}u^{p^\prime }\right) ^{\frac{r}{p^\prime }} \left( \int _{a^{k}}^s\omega _k^q\right) ^{\frac{r}{p}} \omega _k^q(s)ds\right) ^{1/r}:=B_k. \end{aligned}$$

Further we show that

$$\begin{aligned} \sup _{t\in [a^k,a^{k+1}]}\left( \int _{a^{k}}^t\omega _k^qh^{-\frac{p}{q}}\right) ^{\frac{1}{p}} \left( \int _t^{a^{k+2}}u^{p^\prime }\right) ^{\frac{1}{p^\prime }}\lesssim 1. \end{aligned}$$

(3.10)

To this end we write for $t\in [a^k,a^{k+1}]$

$$\begin{aligned}&\int _{a^{k}}^t\omega _k^qh^{-\frac{p}{q}}= \int _{a^{k}}^t\frac{\omega _k^q(x)dx}{\displaystyle \left( \int _x^{a^{k+2}}\left( \int _s^{a^{k+2}}u^{p^\prime }\right) ^{\frac{r}{q^\prime }} \left( \int _{a^{k}}^s\omega _k^q\right) ^{\frac{r}{p}}u^{p^\prime }(s)ds\right) ^{\frac{q}{r}}}\\&\le \left( \int _{a^{k}}^t\omega _k^q\right) \left( \int _t^{a^{k+2}}\left( \int _s^{a^{k+2}}u^{p^\prime }\right) ^{\frac{r}{q^\prime }}\left( \int _{a^{k}}^s\omega _k^q\right) ^{\frac{r}{p}}u^{p^\prime }(s)ds\right) ^{-\frac{p}{r}}\\&\le \left( \int _t^{a^{k+2}}\left( \int _s^{a^{k+2}}u^{p^\prime }\right) ^{\frac{r}{q^\prime }}u^{p^\prime }(s)ds\right) ^{-\frac{p}{r}} \approx \left( \int _t^{a^{k+2}}u^{p^\prime }\right) ^{-\frac{p}{p^\prime }}. \end{aligned}$$

By Hölder’s inequality

$$\begin{aligned}&\Vert T_kf\Vert _{L^q}^q=\int _{a^k}^{a^{k+1}}\omega _k^q(x)\left( \int _x^{a^{k+2}}\log \frac{y-x}{y}f(y)u(y)dy\right) ^qdx\\&\le \left( \int _{a^k}^{a^{k+1}}\omega _k^qh^{r/q}\right) ^{q/r} \left( \int _{a^k}^{a^{k+1}}\omega _k^q(x)h(x)^{-\frac{p}{q}} \left( \int _x^{a^{k+2}}\log \frac{x}{x-y}f(y)u(y)dy\right) ^p dx\right) ^{\frac{q}{p}}\\&\le B^q_k \left( \int _{a^k}^{a^{k+1}}\omega _k^qh^{-\frac{p}{q}} \left( \int _x^{a^{k+2}}\log \frac{x}{y-x}f(y)u(y)dy\right) ^p dx\right) \lesssim B_k^q \Vert f\chi _{[a^k,a^{k+1}]}\Vert _p^q \end{aligned}$$

because of (3.10) and the case $p=q$. Thus, $\Vert T_k\Vert \lesssim B_k$ and

$$\begin{aligned} \Vert T\Vert \lesssim \left( \sum _{k\in {\mathbb {Z}}}\Vert T_k\Vert ^r\right) ^{1/r}\lesssim \left( \sum _{k\in {\mathbb {Z}}}B_k^r\right) ^{1/r}={\mathbb {A}}. \end{aligned}$$

$\square$

Now we estimate the operator $L_3$.

Theorem 3.3

Inequality

$$\begin{aligned} \left( \int _0^\infty w^q(x)\left( \int _{2x}^\infty \log \left( \frac{s-x}{x}\right) \frac{h(s)}{v_1(s)}ds\right) ^qdx \right) ^{1/q}\le C\left( \int _0^\infty h^p\right) ^{1/p},\,\,\,\,h\in {\mathfrak {M}}^+(I) \end{aligned}$$

(3.11)

is equivalent to

$$\begin{aligned} \left( \int _0^\infty \frac{w^q(x)}{x^q}\left( \int _{2x}^{3x}(s-2x) \frac{h(s)}{v_1(s)}ds\right) ^qdx \right) ^{1/q}\le D_1\left( \int _0^\infty h^p\right) ^{1/p} \end{aligned}$$

(3.12)

and

$$\begin{aligned} \left( \int _0^\infty w^q(x)\left( \int _{3x}^\infty \log \left( \frac{s}{x}\right) \frac{h(s)}{v_1(s)}ds\right) ^qdx \right) ^{1/q}\le D_2\left( \int _0^\infty h^p\right) ^{1/p}, \end{aligned}$$

(3.13)

and $C=\Vert L_3\Vert _{L_{v_1}^p(I)\rightarrow L^q_{w}}\approx D_1+D_2$.

Proof

Dividing the interval $(2x,\infty )$ as $(2x,3x]\cup (3x,\infty )$ we decompose (3.11) into two inequalities. Since

$$\begin{aligned} \frac{s-x}{x}=1+\frac{s-2x}{x} \end{aligned}$$

and $\frac{s-2x}{x}<1$, therefore

$$\begin{aligned} \log \frac{s-x}{x}\approx \frac{s-2x}{x}, \,\,s\in [2x,3x] \end{aligned}$$

and the first inequality becomes equivalent to (3.12). For the second inequality we observe that

$$\begin{aligned} \frac{\log 2}{\log 3}\log \left( \frac{s}{x}\right) \le \log \frac{s-x}{x}\le \log \left( \frac{s}{x}\right) , \,\,s\in [3x,\infty ) \end{aligned}$$

and we can replace it equivalently by (3.13). Characterization of (3.12) and (3.13) as well as (3.4) is given in the next Remark. $\square$

Remark 3.4

1. Let $a(x)=3x, k(y,x)=\log \left( \frac{y}{x}\right) .$ Then

$$\begin{aligned} k(y,x)\approx k(y,z)+k(a(z),x),\,\,0<x\le z\le a^{-1}(y)<\infty . \end{aligned}$$

It means that kernel $k(y,x)=\log \left( \frac{y}{x}\right)$ belongs to Oinarov’s class ${\mathbb {O}}_a$ (see [17, Definition 2.3]) and by [17, Corollary 2.2]

$$\begin{aligned} D_2\approx A_0+A_1,\,\,p\le q \end{aligned}$$

and

$$\begin{aligned} D_2\approx B_0+B_1,\,\,q<p, \end{aligned}$$

where

$$\begin{aligned} A_0:=\sup _{0<t<\infty }\left( \int _0^t\log ^q\left( \frac{3t}{x}\right) w^q(x)dx\right) ^{1/q}\left( \int _{3t}^\infty v_1^{-p'}\right) ^{1/p'}, \end{aligned}$$

$$\begin{aligned} A_1:=\sup _{0<t<\infty }\left( \int _0^t w^q\right) ^{1/q}\left( \int _{3t}^\infty \log ^{p'}\left( \frac{y}{t}\right) v_1^{-p'}(y)dy\right) ^{1/p'}, \end{aligned}$$

$$\begin{aligned} B_0:=\left( \left( \int _0^{t/3}\log ^q\left( \frac{t}{x}\right) w^q(x)dx\right) ^{r/q}\left( \int _{t}^\infty v_1^{-p'}\right) ^{r/q'}v_1^{-p'}(t)dt\right) ^{1/r}, \end{aligned}$$

$$\begin{aligned} B_1:=\left( \left( \int _0^{t}w^q\right) ^{r/p}\left( \int _{3t}^\infty \log ^{p'}\left( \frac{y}{t}\right) v_1^{-p'}(y)dy\right) ^{r/p'}w^q(t)dt\right) ^{1/r}. \end{aligned}$$

2. Inequalities (3.4) and (3.12) are characterized by [17, Theorem 3.2] and [17, Theorem 3.1], respectively. We formulate the results for the case $p\le q.$ The case $q<p$ is also characterized, but in discrete form and we omit details.

Applying [17, Theorem 3.2] for the best constant D in (3.4) we find

$$\begin{aligned} D\approx {{\mathcal {A}}}_0+{{\mathcal {A}}}_1,\,\,p\le q, \end{aligned}$$

where

$$\begin{aligned} {{\mathcal {A}}}_0:=\sup _{0<t<\infty }\sup _{\frac{3t}{2}<s<t}\left( \int _s^t(x-s)^q w^q(x)dx\right) ^{1/q}\left( \int _{\frac{3t}{2}}^{2s} v_1^{-p'}\right) ^{1/p'},\\ {{\mathcal {A}}}_1:=\sup _{0<t<\infty }\sup _{\frac{3t}{2}<s<t}\left( \int _s^t w^q\right) ^{1/q}\left( \int _{\frac{3t}{2}}^{2s}(2s-y)^{p'}v_1^{-p'}(y)dy\right) ^{1/p'}. \end{aligned}$$

Analogously, applying [17, Theorem 3.1] for the best constant $D_1$ in (3.12) we obtain

$$\begin{aligned} D_1\approx {{\mathcal {A}}}^*_0+{{\mathcal {A}}}^*_1,\,\,p\le q, \end{aligned}$$

where

$$\begin{aligned} {{\mathcal {A}}}^*_0:=\sup _{0<t<\infty }\sup _{s<t<\frac{3s}{2}}\left( \int _s^t(t-x)^q w^q(x)dx\right) ^{1/q}\left( \int _{2t}^{3s} v_1^{-p'}\right) ^{1/p'},\\ {{\mathcal {A}}}^*_1:=\sup _{0<t<\infty }\sup _{s<t<\frac{3s}{2}}\left( \int _s^t w^q\right) ^{1/q}\left( \int _{2t}^{3s}(y-2t)^{p'}v_1^{-p'}(y)dy\right) ^{1/p'}. \end{aligned}$$

Example 3.5

Let $1/p'<\alpha <1+1/p',$

$$\begin{aligned} W^1_{p,\alpha }(I):=\{f\in AC(I):\Vert f\Vert _{W^1_{p,\alpha }}:=\Vert f\Vert _{L^p_{x^{\alpha -1}}}+\Vert f^\prime \Vert _{L^p_{x^{\alpha }}}<\infty \}. \end{aligned}$$

Then

$$\begin{aligned} \Vert Hf\Vert _{L^p_{x^{\alpha }}}\lesssim \Vert f\Vert _{W^1_{p,\alpha }}. \end{aligned}$$

(3.14)

We have

$$\begin{aligned} \Vert f\Vert _{W^1_{p,\alpha }}\approx \Vert f^\prime \Vert _{L^p_{x^{\alpha }}} \end{aligned}$$

and by the upper bound similar to (2.7) we write

$$\begin{aligned} \Vert H\Vert _{W^1_{p,\alpha }(I)\rightarrow L^p_{x^{\alpha }}(I)}\lesssim \sum _{i=1}^3 \Vert L_i\Vert _{L^p_{x^{\alpha }}(I)\rightarrow L^p_{x^{\alpha }}(I)}. \end{aligned}$$

(3.15)

Observe that

$$\begin{aligned} L_1g(x)+L_2g(x)=\int _0^{2x}\log \left( \frac{x}{|s-x|}\right) g(s)ds \end{aligned}$$

and

$$\begin{aligned} \Vert L_1g+L_2g\Vert _{L^p_{x^{\alpha }}}\lesssim \Vert g\Vert _{L^p_{x^{\alpha }}} \end{aligned}$$

is equivalent to

$$\begin{aligned} J:=\left( \int _0^\infty \left( \int _0^{2x}\log \left( \frac{x}{|s-x|}\right) \frac{h(s)}{s^\alpha }ds\right) ^p x^{\alpha p}dx \right) ^{1/p}\lesssim \left( \int _0^\infty h^p\right) ^{1/p} \end{aligned}$$

(3.16)

for all $h\in {\mathfrak {M}}^+(I).$ By change of variables and applying Minkowskii’s inequality we obtain

$$\begin{aligned} J&=\left( \int _0^\infty \left( \int _0^{2}\log \left( \frac{1}{|t-1|}\right) \frac{h(tx)}{t^\alpha }dt\right) ^p x^pdx \right) ^{1/p}\\&\le \int _0^{2}t^{-\alpha }\log \left( \frac{1}{|t-1|}\right) \left( \int _0^\infty h^p(tx)x^pdx\right) ^{1/p}\\&=\int _0^{2}t^{-\alpha -1/p}\log \left( \frac{1}{|t-1|}\right) dt\Vert h\Vert _{L^p}=:\Lambda \Vert h\Vert _{L^p}. \end{aligned}$$

Let $\Lambda =\Lambda _1+\Lambda _2,$ where

$$\begin{aligned} \Lambda _2\approx \int _1^2\log \left( \frac{1}{|t-1|}\right) dt=-\int _0^1\log tdt<\infty \end{aligned}$$

and

$$\begin{aligned} \Lambda _1=-\int _0^1 t^{-\alpha -1/p}\log (1-t)dt. \end{aligned}$$

Since

$$\begin{aligned} -\log (1-t)\approx t\chi _{(0,1/2)}(t)-\chi _{(1/2,1)}(t)\log (1-t), \end{aligned}$$

then

$$\begin{aligned} \Lambda _1\approx \int _0^{1/2}t^{\frac{1}{p'}-\alpha }dt-\int _{1/2}^1\log (1-t)dt<\infty \end{aligned}$$

provided $\alpha <1+\frac{1}{p'}$ and (3.16) follows.

The proof of $\Vert L_3g\Vert _{L^p_{x^{\alpha }}}\lesssim \Vert g\Vert _{L^p_{x^{\alpha }}}$ is similar.

4 Main Result

Theorem 4.1

Let $1<p, q<\infty .$ Suppose that ${{\mathbf {A}}}_1<\infty$ (see (2.3)), $v_1\in {\mathfrak {M}}^\uparrow (I)$ and there is $\gamma >0$ such that $x^{-\gamma }v_1(x)\in {\mathfrak {M}}^\downarrow (I).$ Then

(4.1)

where ${{\mathbb {D}}}=A$ (see (3.1)) if $p\le q$ and ${{\mathbb {D}}}=B$ (see (3.2)) if $q<p$ and the constants ${{\mathbb {A}}}, D, D_1, D_2$ are determined by (3.5), (3.4), (3.12), (3.13), respectively. Explicit values of $D, D_1, D_2$ are described in Remark 3.4.

Proof

It follows from (2.7) and Theorems 3.1, 3.2, 3.3. $\square$